James, Thank you for your help.
I appreciate the very thorough explanation. I have never heard of "noise" although I have produced a 'kicked' map in phenix and refined the structure using those maps..though I guess this is different. I will try it. Thanks. -Yarrow > > Formally, the "best" way to compare B factors in two structures with > different average B is to add a constant to all the B factors in the > low-B structure until the average B factor is the same in both > structures. Then you can compare "apples to apples" as it were. The > "extra B" being added is equivalent to "blurring" the more well-ordered > map to make it match the less-ordered one. Subtracting a B factor from > the less-ordered structure is "sharpening", and the reason why you > shouldn't do that here is because you'd be assuming that a sharpened map > has just as much structural information as the better diffracting > crystal, and that's obviously no true (not as many spots). In reality, > your comparison will always be limited by the worst-resolution data you > have. > > Another reason to add rather than subtract a B factor is because B > factors are not really "linear" with anything sensible. Yes, B=50 is > "more disordered" than B=25, but is it "twice as disordered"? That > depends on what you mean by "disorder", but no matter how you look at > it, the answer is generally "no". > > One way to define the "degree of disorder" is the volume swept out by > the atom's nucleus as it "vibrates" (or otherwise varies from cell to > cell). This is NOT proportional to the B-factor, but rather the 3/2 > power of the B factor. Yes, 3/2 power. The value of "B", is > proportional to the SQUARE of the width of the probability distribution > of the nucleus, so to get the volume of space swept out by it you have > to take the square root to get something proportional the the width and > then you take the 3rd power to get something proportional to the volume. > > An then, of course, if you want to talk about the electron cloud (which > is what x-rays "see") and not the nuclear position (which you can only > see if you are a neutron person), then you have to "add" a B factor of > about 8 to every atom to account for the intrinsic width of the electron > cloud. Formally, the B factor is "convoluted" with the intrinsic atomic > form factor, but a "native" B factor of 8 is pretty close for most atoms. > > For those of you who are interested in something more exact than > "proportional" the equation for the nuclear probability distribution > generated by a given B factor is: > kernel_B(r) = (4*pi/B)^1.5*exp(-4*pi^2/B*r^2) > where "r" is the distance from the "average position" (aka the x-y-z > coordinates in the PDB file). Note that the width of this distribution > of atomic positions is not really an "error bar", it is a "range". > There's a difference between an atom actually being located in a variety > of places vs not knowing the centroid of all these locations. Remember, > you're averaging over trillions of unit cells. If you collect a > different dataset from a similar crystal and re-refine the structure the > final x-y-z coordinate assigned to the atom will not change all that much. > > The full-width at half-maximum (FWHM) of this kernel_B distribution is: > fwhm = 0.1325*sqrt(B) > and the probability of finding the nucleus within this radius is > actually only about 29%. The radius that contains the nucleus half the > time is about 1.3 times wider, or: > r_half = 0.1731*sqrt(B) > > That is, for B=25, the atomic nucleus is within 0.87 A of its average > position 50% of the time (a volume of 2.7 A^3). Whereas for B=50, it is > within 1.22 A 50% of the time (7.7 A^3). Note that although B=50 is > twice as big as B=25, the half-occupancy radius 0.87 A is not half as > big as 1.22 A, nor are the volumes 2.7 and 7.7 A^3 related by a factor > of two. > > Why is this important for comparing two structures? Since the B factor > is non-linear with disorder, it is important to have a common reference > point when comparing them. If the low-B structure has two atoms with > B=10 and B=15 with average overall B=12, that might seem to be > "significant" (almost a factor of two in the half-occupancy volume) but > if the other structure has an average B factor of 80, then suddenly 78 > vs 83 doesn't seem all that different (only a 10% change). Basically, a > difference that would be "significant" in a high-resolution structure is > "washed out" by the overall crystallographic B factor of the > low-resolution structure in this case. > > Whether or not a 10% difference is "significant" depends on how accurate > you think your B factors are. If you "kick" your coordinates (aka using > "noise" in PDBSET) and re-refine, how much do the final B factors change? > > -James Holton > MAD Scientist > > On 2/25/2013 12:08 PM, Yarrow Madrona wrote: >> Hello, >> >> Does anyone know a good method to compare B-factors between structures? >> I >> would like to compare mutants to a wild-type structure. >> >> For example, structure2 has a higher B-factor for residue X but how can >> I >> show that this is significant if the average B-factor is also higher? >> Thank you for your help. >> >> > > -- Yarrow Madrona Graduate Student Molecular Biology and Biochemistry Dept. University of California, Irvine Natural Sciences I, Rm 2403 Irvine, CA 92697
